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It is well known that the area of a torus $T^{2}=S^{1}\times S^{1}$ is the product of the circumferences of the circles $S^{1}$.

Is the surface area of a torus $T^{n}$ generalizable in the form of the product $\prod_{i}^{n}\pi d_{i}$?

Where $d_{i}$is the circumference of the $i$-th circle and $T^{n}=\underbrace{S^{1}\times \ldots \times S^{1}}_{n}$.

Kurious
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  • When $n=3$ and the $d_i$s are measured in meters, your formula gives something in cubic meters as the area!? Houston, we have a problem! – Jyrki Lahtonen Jul 25 '23 at 05:14
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    Of course, your observation is correct. What I had in mind when I asked the question is: Since an n-torus is a hypertorus, how do you calculate its hypersurface? In the same sense that the surface area of a 4-sphere goes as the cube of its radius. – Kurious Jul 25 '23 at 05:19
  • Assuming you really are asking about its (hyper)volume, I guess you get something definite, if you also specify an embedding of $T^n$ into some Euclidean space. I'm relatively ignorant about this, but the choice of embedding may make a difference in general. – Jyrki Lahtonen Jul 25 '23 at 05:21
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    @Kurious a language issue, but nevertheless: the surface area of a $4$-sphere goes as $r^4$. Note, an $n$-sphere is an $n$-dimensional object (even if one often defines it as a subset of $\Bbb{R}^{n+1}$). So, $1$-sphere, $S^1$ has “1-volume” (i.e length) proportional to the radius ($2\pi r$), while $S^2$ has “2-volume” which grows as $r^2$ (specifically $4\pi r^2$) and so on. – peek-a-boo Jul 25 '23 at 06:41

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In general suppose you have two Riemannian manifolds $(M_1,g_1)$ and $(M_2,g_2)$, and you consider the Riemannian product manifold $M=M_1\times M_2$ with Riemannian metric $g=g_1\oplus g_2$ (i.e for all $(x,y)\in M_1\times M_2$ and tangent vectors $(\xi,\eta)\in T_xM_1\times T_yM_2\cong T_{(x,y)}(M_1\times M_2)$, $g(\xi,\eta):=g_1(\xi)+g_2(\eta)$… or more abstractly, if $\pi_i:M\to M_i$ is the projection, then $g:=\pi_1^*g_1+\pi_2^*g_2$).

If you let $\lambda_g,\lambda_{g_1},\lambda_{g_2}$ (often denoted as $dV_g,dV_{g_1},dV_{g_2}$ or simply $dV_M,dV_{M_1},dV_{M_2}$ if the metric is understood) be the induced measures on $M,M_1,M_2$ respectively, then you can prove that (say on the Borel $\sigma$-algebra) $\lambda_g=\lambda_{g_1}\times\lambda_{g_2}$ is the product measure. Or what amounts to the same thing in the oriented case, if $\omega_g,\omega_{g_1},\omega_{g_2}$ are the corresponding Riemannian volume forms (which also are commonly denoted by the corresponding $dV$ notation), then $\omega_g=(\pi_1^*\omega_{g_1})\wedge (\pi_2^*\omega_{g_2})$. Hence, $\lambda_g(M\times N)=\lambda_{g_1}(M_1)\cdot \lambda_{g_2}(M_2)$.

If we now apply this $n$-times with a torus of “multi-radius” $r=(r_1,\dots, r_n)$, i.e $T^n(r):=S^1(r_1)\times \dots, \times S^1(r_n)$, where $S^1(r_i)$ denotes the usual circle of radius $r_i$, then we find that the $n$-dimensional measure (i.e hypervolume) of $T^n(r)$ is $\prod_{i=1}^n2\pi r_i=(2\pi)^nr_1\cdots r_n$. So indeed, if you let $d_i=2r_i$ be the diameter, then this is simply $\prod_{i=1}^n\pi d_i=\pi^nd_1\cdots d_n$. Of course, I have equipped the circle with its usual metric, and the torus with the corresponding product metric.

peek-a-boo
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